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Difference between revisions of "2015 AIME II Problems/Problem 7"

(→‎Solution #2)
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Now consider the rectangle <math>PQRS</math>. Since <math>SR</math> is collinear with <math>BC</math> and parallel to <math>PQ</math>, <math>PQ</math> is parallel to <math>BC</math> meaning <math>\Delta APQ</math> is similar to <math>\Delta ABC</math>.  
 
Now consider the rectangle <math>PQRS</math>. Since <math>SR</math> is collinear with <math>BC</math> and parallel to <math>PQ</math>, <math>PQ</math> is parallel to <math>BC</math> meaning <math>\Delta APQ</math> is similar to <math>\Delta ABC</math>.  
 
Let <math>F</math> be the intersection between <math>AE</math> and <math>PQ</math>. By the similar triangles, we know that <math>\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}</math>. Since <math>PF+FQ=PQ=\omega</math>. We can solve for <math>PF</math> and <math>FQ</math> in terms of <math>\omega</math>. We get that <math>PF=\frac{48}{125} \omega</math> and <math>FQ=\frac{77}{125} \omega</math>.
 
Let <math>F</math> be the intersection between <math>AE</math> and <math>PQ</math>. By the similar triangles, we know that <math>\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}</math>. Since <math>PF+FQ=PQ=\omega</math>. We can solve for <math>PF</math> and <math>FQ</math> in terms of <math>\omega</math>. We get that <math>PF=\frac{48}{125} \omega</math> and <math>FQ=\frac{77}{125} \omega</math>.
 +
Let's work with <math>PF</math>. We know that <math>PQ</math> is parallel to <math>BC</math> so <math>\Delta APF</math> is similar to <math>\Delta ABE</math>. We can set up the proportion:
 +
<math>\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}</math>. Solving for <math>AF</math>, <math>AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega</math>.
 +
We can solve for <math>PS</math> then since we know that <math>PS=FE</math> and <math>FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega</math>.
 +
Therefore, <math>[PQRS] = PQ \cdot PS = \omega ()</math>
  
 
==See also==
 
==See also==
 
{{AIME box|year=2015|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2015|n=II|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:35, 29 March 2015

Problem

Triangle $ABC$ has side lengths $AB = 12$, $BC = 25$, and $CA = 17$. Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$, vertex $Q$ on $\overline{AC}$, and vertices $R$ and $S$ on $\overline{BC}$. In terms of the side length $PQ = w$, the area of $PQRS$ can be expressed as the quadratic polynomial

Area($PQRS$) = $\alpha w - \beta \cdot w^2$.

Then the coefficient $\beta = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

If $\omega = 25$, the area of rectangle $PQRS$ is $0$, so

\[\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0\]

and $\alpha = 25\beta$. If $\omega = \frac{25}{2}$, we can reflect $APQ$ over PQ, $PBS$ over $PS$, and $QCR$ over $QR$ to completely cover rectangle $PQRS$, so the area of $PQRS$ is half the area of the triangle. Using Heron's formula, since $s = \frac{12 + 17 + 25}{2} = 27$,

\[[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90\]

so

\[45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}\]

and

\[\beta = \frac{180}{625} = \frac{36}{125}\]

so the answer is $m + n = 36 + 125 = \boxed{161}$.

Solution #2

Diagram: (Diagram goes here)

Similar triangles can also solve the problem. First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $BC$ and solving. (The $8$ side would be collinear with line $AB$) After finding the area, solve for the altitude to $BC$. Let $E$ be the intersection of the altitude from $A$ and side $BC$. Then $AE = \frac{36}{5}$. Solving for $BE$ using the Pythagorean Formula, we get $BE = \frac{48}{5}$. We then know that $CE = \frac{77}{5}$. Now consider the rectangle $PQRS$. Since $SR$ is collinear with $BC$ and parallel to $PQ$, $PQ$ is parallel to $BC$ meaning $\Delta APQ$ is similar to $\Delta ABC$. Let $F$ be the intersection between $AE$ and $PQ$. By the similar triangles, we know that $\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}$. Since $PF+FQ=PQ=\omega$. We can solve for $PF$ and $FQ$ in terms of $\omega$. We get that $PF=\frac{48}{125} \omega$ and $FQ=\frac{77}{125} \omega$. Let's work with $PF$. We know that $PQ$ is parallel to $BC$ so $\Delta APF$ is similar to $\Delta ABE$. We can set up the proportion: $\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}$. Solving for $AF$, $AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega$. We can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega$. Therefore, $[PQRS] = PQ \cdot PS = \omega ()$

See also

2015 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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